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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 15 8β’2 Lesson 15: Informal Proof of the Pythagorean Theorem Student Outcomes ο§ Students are introduced to the Pythagorean theorem and are shown an informal proof of the theorem. ο§ Students use the Pythagorean theorem to find the length of the hypotenuse of a right triangle. Lesson Notes Since 8.G.B.6 and 8.G.B.7 are post-test standards, this lesson is designated as an extension lesson for this module. However, the content within this lesson is prerequisite knowledge for Module 7. If this lesson is not used with students as part of the work within Module 2, it must be used with students prior to beginning work on Module 7. Please realize that many mathematicians agree that the Pythagorean theorem is the most important theorem in geometry and has immense implications in much of high school mathematics in general (e.g., when studying quadratics and trigonometry). It is crucial that students see the teacher explain several proofs of the Pythagorean theorem and practice using it before being expected to produce a proof on their own. Classwork Concept Development (5 minutes) The Pythagorean theorem is a famous theorem that is used throughout much of high school mathematics. For that reason, students see several proofs of the theorem throughout the year and have plenty of practice using it. The first thing to know about the Pythagorean theorem is what it states. PYTHAGOREAN THEOREM: If the lengths of the legs of a right triangle are π and π, and the length of the hypotenuse is π, then π2 + π 2 = π 2 . Given a right triangle π΄π΅πΆ with πΆ being the vertex of the right angle, then the sides Μ Μ Μ Μ π΄πΆ and Μ Μ Μ Μ π΅πΆ are called the legs of Μ Μ Μ Μ β³ π΄π΅πΆ, and π΄π΅ is called the hypotenuse of β³ π΄π΅πΆ. Scaffolding: Draw arrows, one at a time, to show that each side is the opposite of the given angle. Take note of the fact that side π is opposite the angle π΄, side π is opposite the angle π΅, and side π is opposite the angle πΆ. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 180 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 Discussion (15 minutes) The first proof of the Pythagorean theorem requires knowledge of some basic facts about geometry. 1. Congruent triangles have equal areas. 2. All corresponding parts of congruent triangles are congruent. 3. The triangle sum theorem. (β sum of β³) a. For right triangles, the two angles that are not the right angle have a sum of 90°. (β sum of rt. β³) What we look at next is what is called a square within a square. The outside square has side lengths (π + π), and the inside square has side lengths π. Our goal is to show that π2 + π 2 = π 2 . To accomplish this goal, we compare the total area of the outside square with the parts it is composed of, that is, the four triangles and the smaller inside square. Note to Teacher: Remind students to use the distributive law to determine the area of the outside square. Also remind them to use what they know about exponential notation to simplify the expression. Ask students the following questions during the discussion: ο§ Looking at the outside square only, the square with side lengths (π + π), what is its area? οΊ ο§ Are the four triangles with sides lengths π and π congruent? If so, how do you know? οΊ ο§ Yes, the triangles are congruent. From the diagram, we can see that each triangle has a right angle, and each triangle has side lengths of π and π. Our rigid motions preserve those measures, and we can trace one triangle and use rigid motions to prove that they are congruent. What is the area of just one triangle? οΊ ο§ The area of the outside square is (π + π)2 which after expanding is π2 + 2ππ + π 2. 1 2 ππ Does each triangle have the same area? If so, what is the sum of all four of those areas? οΊ Yes, each triangle has the same area because they are congruent. The sum of all four triangles is 1 2 4 ( ππ) or after simplifying, 2ππ. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 181 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM ο§ For right triangles, we know that one angle has a measure of 90°. Since the sum of the measures of all three angles must be 180°, we know that the other two angle measures must have a sum of 90°. Now, look at just one side of the figure. We have an angle with a red arc and an angle with a blue arc. In between them is another angle that we do not know the measure of. All three angles added together make up the straight side of the outside figure. What must be the measure of the unknown angle (the measure of the angle between the red and blue arcs)? How do you know? οΊ ο§ The sum of the measures of the interior angles of a triangle is 180°. What is the sum of the measures of the two interior angles of a right triangle, not including the right angle (β sum of rt. β³)? How do you know? οΊ ο§ Corresponding angles of congruent triangles are also congruent and equal in measure. So we know that the angles marked by the red arcs are equal in measure, and the angles marked with the blue arcs and line are equal in measure. What do we know about the sum of the interior angles of a triangle (β sum of β³)? οΊ ο§ 8β’2 We called this entire figure a square within a square, but we want to make sure that the figure in the center is indeed a square. To do so, we need to look at the angles of the triangles. First of all, what do we know about corresponding angles of congruent triangles? οΊ ο§ Lesson 15 Since the angle with the red arc and the angle with the blue arc must have a sum of 90°, and all three angles together must make a straight angle measuring 180° (β s on a line), the unknown angle must equal 90°. That means that the figure with side lengths π must be a square. It is a figure with four equal sides and four right angles. What is the area of this square? οΊ The area of the square must be π 2 . Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 182 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8β’2 Recall our goal: To show that π2 + π 2 = π 2 . To accomplish this goal, we compare the total area of the outside square with the parts it is composed of, that is, the four triangles and the smaller, inside square. Do we have everything we need to accomplish our goal? οΊ Yes, we know the area of the outside square is π2 + 2ππ + π 2; the sum of the areas of the four triangles is 2ππ, and the area of the inside square, π 2 . Show students the end of the square within a square proof: Total area of the outside square = area of four triangles + area of inside square π2 + 2ππ + π 2 = 2ππ + π 2 π + 2ππ β 2ππ + π 2 = 2ππ β 2ππ + π 2 π2 + π 2 = π 2 2 Thus, we have shown the Pythagorean theorem to be true using a square within a square. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 183 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 Example 1 (2 minutes) Example 1 Now that we know what the Pythagorean theorem is, letβs practice using it to find the length of a hypotenuse of a right triangle. Determine the length of the hypotenuse of the right triangle. The Pythagorean theorem states that for right triangles ππ + ππ = ππ , where π and π are the legs, and π is the hypotenuse. Then, π π + π π = ππ ππ + ππ = ππ ππ + ππ = ππ πππ = ππ . Since we know that πππ = πππ , we can say that the hypotenuse π is ππ. Example 2 (3 minutes) Example 2 Determine the length of the hypotenuse of the right triangle. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 184 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM ο§ Based on our work in the last example, what should we do to find the length of the hypotenuse? οΊ ο§ 8β’2 Use the Pythagorean theorem, and replace π and π with 3 and 7. Then, π2 + π 2 = π 2 32 + 72 = π 2 9 + 49 = π 2 58 = π 2 . Since we do not know what number times itself produces 58, for now we can leave our answer as 58 = π 2 . Later this year, we learn how to determine the actual value for π for problems like this one. Exercises 1β5 (10 minutes) Exercises 1β5 For each of the exercises, determine the length of the hypotenuse of the right triangle shown. Note: Figures are not drawn to scale. 1. π π + π π = ππ ππ + ππ = ππ π + ππ = ππ ππ = ππ π=π 2. π π + π π = ππ ππ + πππ = ππ ππ + πππ = ππ πππ = ππ 3. π π + π π = ππ ππ + ππ = ππ ππ + ππ = ππ ππ = ππ Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 185 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 4. π π + π π = ππ ππ + ππ = ππ π + ππ = ππ ππ = ππ 5. π π + π π = ππ ππ + ππ = ππ π + ππ = ππ ππ = ππ Closing (5 minutes) Summarize, or have students summarize, the lesson. ο§ We were shown a proof for the Pythagorean theorem that required us to find the area of four congruent triangles and two squares. ο§ We learned that right triangles have sides π and π, known as legs, and a side π, known as the hypotenuse. ο§ We know that for right triangles, π2 + π 2 = π 2 . ο§ We learned how to use the Pythagorean theorem to find the length of the hypotenuse of a right triangle. Lesson Summary Μ Μ Μ Μ and π©πͺ Μ Μ Μ Μ are called the legs Given a right triangle π¨π©πͺ with πͺ being the vertex of the right angle, then the sides π¨πͺ Μ Μ Μ Μ is called the hypotenuse of β³ π¨π©πͺ. of β³ π¨π©πͺ, and π¨π© Take note of the fact that side π is opposite the angle π¨, side π is opposite the angle π©, and side π is opposite the angle πͺ. The Pythagorean theorem states that for any right triangle, ππ + ππ = ππ . Exit Ticket (5 minutes) Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 186 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM Name 8β’2 Date Lesson 15: Informal Proof of the Pythagorean Theorem Exit Ticket 1. Label the sides of the right triangle with leg, leg, and hypotenuse. 2. Determine the length of π in the triangle shown. 3. Determine the length of π in the triangle shown. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 187 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 Exit Ticket Sample Solutions 1. Label the sides of the right triangle with leg, leg, and hypotenuse. hypotenuse leg leg 2. Determine the length of π in the triangle shown. π π + π π = ππ ππ + ππ = ππ ππ + ππ = ππ πππ = ππ ππ = π 3. Determine the length of π in the triangle shown. π π + π π = ππ ππ + ππ = ππ ππ + ππ = ππ ππ = ππ Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 188 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 Problem Set Sample Solutions Students practice using the Pythagorean theorem to find the length of the hypotenuse of a right triangle. For each of the problems below, determine the length of the hypotenuse of the right triangle shown. Note: Figures are not drawn to scale. 1. π π + π π = ππ ππ + ππ = ππ π + ππ = ππ ππ = ππ 2. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ 3. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ 4. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 189 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 5. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ π=π 6. 7. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ π π + π π = ππ πππ + ππ = ππ πππ + ππ = ππ πππ = ππ ππ = π 8. Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 190 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 15 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’2 9. π π + π π = ππ πππ + ππ = ππ πππ + ππ = ππ πππ = ππ 10. π π + π π = ππ πππ + ππ = ππ πππ + ππ = ππ πππ = ππ 11. π π + π π = ππ πππ + ππ = ππ πππ + ππ = ππ πππ = ππ ππ = π 12. π π + π π = ππ ππ + ππ = ππ ππ + π = ππ ππ = ππ Lesson 15: Informal Proof of the Pythagorean Theorem This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 191 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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